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4 years ago

If F(x) = 3x-2and g(x) = 2x+ 1,find (f-g)(x)

2 answers:

6 0

(F)3x-2(g)2x+1
Your subtracting to subtract the two x
(3x-2)-(2x+1)
The parenthesis is to know they are separate and not a negative number not subtract
ANSWER:x-3

sattari [20]4 years ago

3 0

Answer:

Step-by-step explanation:

(f - g)(x) = f(x) - g(x)

We have f(x) = 3x - 2 and g(x) = 2x + 1. Substitute:

(f - g)(x) = (3x - 2) - (2x + 1)

(f - g)(x) = 3x - 2 - 2x - 1 <em>combine like terms</em>

(f - g)(x) = (3x - 2x) + (-2 - 1)

(f - g)(x) = x - 3

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I am part of a whole. I am greater than three fourths but less than nine tenths. I am a decimal with a 3 in its hundredth place.

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Lets get started :)

Given is that the value is greater than three fourths { $\frac{3}{4}$ = 0.75 }
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Fractions between $\frac{3}{4}$ and $\frac{9}{10}$ are
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3 years ago

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The data (1.5.8.5.1) represent a random sample of the number of days absent from school for five students at Monta Vista High. F

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Answer:

$\bar X = \frac{1+5+8+5+1}{5}= \frac{20}{5}= 4$

$\sigma= \sqrt{\frac{(1-4)^2 +(5-4)^2 +(8-4)^2 +(5-4)^2 +(1-4)^2}{5}} =\sqrt{\frac{36}{5}}= 2.68$

And based on this the best answer would be:

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Step-by-step explanation:

For this case we have the following data given:

1,5,8,5,1

We can find the sample mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And replacing we got:

$\bar X = \frac{1+5+8+5+1}{5}= \frac{20}{5}= 4$

And for the deviation (assuming that the correct approximation is the deviation for a population) we can calculate the deviation with the following formula:

$\sigma = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n}}$

And replacing we got:

$\sigma= \sqrt{\frac{(1-4)^2 +(5-4)^2 +(8-4)^2 +(5-4)^2 +(1-4)^2}{5}} =\sqrt{\frac{36}{5}}= 2.68$

And based on this the best answer would be:

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